Answers To Basic Mensa Logic Problems

These refer back to the original problems in my June 25 post:

• Three Basic Mensa Logic Problems
• 
  

1)This is straightforward. Obviously, a consecutive clockwise and counter-clockwise rotation of 60 degrees cancel each other. Then, rotating the triangle through 360 degrees clockwise simply turns it through a full circle - bringing it to its original position. Thus, the sketch:

replicates: (A), i.e. the same as the original orientation.

(2) This references a series of operations on a square whose top left corner is A, and the other corners (going clockwise) are: B, C and D. The operation sequence is: R2 M1 M3, where:

R2: rotation clockwise by 180 degrees

M1: Mirrors corners exactly through the midline of the square

M3: Mirrors two opposing corners in the opposite sense to M2

Performing operation R2 first causes the square to rotate so that 'A' ends up where C is in the original orientation. (Fig. 1(A)) Operation M1 then causes all the existing corners to be mirrored through the midline of the square. Thus, Fig 1 (B) shows the orientation after M1.

The last operation, M3, mirrors two opposing diagonal corners: from top right to bottom left (e.g. opposite to the sense of M2) This will yield the result shown in Fig. 2 which conforms to a representation DABC.

(3) The next step in the sequence is easy to figure out once one identifies the key components of the previous sequence of 4 and how they are changing. This is in the direction of increasing white squares and reducing the orange. Since the algorithmic sequence requires at least two whites for the next step, this automatically eliminates options (A), (C) and (D), leaving (B).